Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

\(500 \mathrm{~mL}\) of a hydrocarbon gas burnt in excess of oxygen, gave \(2500 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) and \(3000 \mathrm{~mL}\) of water vapour (at constant T and P). The formula of the hydrocarbon is (a) \(\mathrm{C}_{3} \mathrm{H}_{6}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{s} \mathrm{H}_{12}\) (d) \(\mathrm{CH}_{4}\)

Step-by-Step Solution

Answer

The formula of the hydrocarbon is (c) \(\mathrm{C}_5\mathrm{H}_{12}\).

Step 1: Understanding the Reaction

When a hydrocarbon gas burns in excess oxygen, it undergoes a combustion reaction producing carbon dioxide and water. The general formula for such a reaction is: \[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \] We need to determine the values of \(x\) and \(y\) that correspond to the given products.

Step 2: Analyze Volume of Products

From the problem statement, burning 500 mL of the hydrocarbon produced 2500 mL of \(CO_2\) and 3000 mL of \(H_2O\). Using the idea that gases under the same conditions of temperature and pressure have the same volume proportions as their mole ratios, the coefficients in the balanced chemical equation directly correspond to volumes.

Step 3: Writing Volume Ratios

From the reaction, we know: - 2500 mL of \(CO_2\) suggests 5 times the volume of the hydrocarbon.- 3000 mL of \(H_2O\) suggests 6 times the volume of the hydrocarbon.This suggests the ratio of carbon dioxide to water vapor to reactant is 5:6:1.

Step 4: Determining the Hydrocarbon Formula

Using the balanced reaction's mole ratio: - Each mole of carbon in the hydrocarbon gives one mole of \(CO_2\), thus \(x = 5\).- Each pair of moles of hydrogen in the hydrocarbon gives one mole of \(H_2O\), so \(\frac{y}{2} = 6\) thus \(y = 12\). Therefore, the formula of the hydrocarbon is \(\mathrm{C}_5\mathrm{H}_{12}\).

Step 5: Select the Correct Option

Compare the calculated formula \(\mathrm{C}_5\mathrm{H}_{12}\) with the given options. Option (c) \(\mathrm{C}_5\mathrm{H}_{12}\) matches this formula exactly.

Key Concepts

Volume Ratios in Reactions

When analyzing chemical reactions involving gases, we can use the concept of volume ratios to determine the relationships between reactants and products. Because gases at constant temperature and pressure occupy volumes proportional to the number of moles, the coefficients from a balanced chemical equation can be converted directly into volume ratios.
In the case of hydrocarbon combustion, the reaction typically involves a hydrocarbon and oxygen reacting to form carbon dioxide and water.

• Volume ratios tell us how much of each gas is produced or consumed relative to another.
• For instance, if we know that 500 mL of a hydrocarbon produce 2500 mL of carbon dioxide and 3000 mL of water vapor, we can determine a ratio of 1:5:6.
• This means for every 1 volume of hydrocarbon, 5 volumes of carbon dioxide and 6 volumes of water vapor are produced.

Understanding these ratios allows us to further explore the balanced reaction and molar ratios of the compounds involved.

Balanced Chemical Equation

A balanced chemical equation is essential to accurately describe the proportions of reactants and products in a chemical reaction. Balancing an equation means ensuring that the number of atoms for each element is the same on both sides of the equation. This reflects the conservation of mass, as atoms are neither created nor destroyed within a chemical reaction.
The simplified form of a combustion reaction of a hydrocarbon is: \[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \]

• The coefficients before each compound indicate the number of moles, and thus volumes, involved in the reaction.
• For example, one mole of a hydrocarbon might lead to the production of a specific number of moles of carbon dioxide and water.
• In our exercise, the balanced equation for the hydrocarbon C\(_5\)H\(_{12}\) is adjusted so that it shows that 5 volumes of CO\(_2\) and 6 volumes of H\(_2\)O are produced from each volume of C\(_5\)H\(_{12}\).

This balance provides clarity to the stoichiometry and allows us to accurately describe volumetric relationships.

Determining Molecular Formula

To determine the molecular formula of a compound from a reaction, we must analyze the volume ratios and relate them to the stoichiometry of the reaction. From the volumes of the products in the given problem (2500 mL of CO\(_2\) and 3000 mL of H\(_2\)O), we can deduce the formula of the reacting hydrocarbon.

• Each carbon atom in the hydrocarbon yields one molecule of CO\(_2\).
• The 2500 mL of CO\(_2\) suggests 5 atoms of carbon, as the volume ratio from the problem gives us a 5:1 ratio with the hydrocarbon.
• Similarly, two hydrogen atoms form one molecule of H\(_2\)O, requiring 12 hydrogen atoms to match the 3000 mL of H\(_2\)O at a 6:1 ratio with the hydrocarbon.
• This leads to a hydrocarbon formula of C\(_5\)H\(_{12}\), matching one of the given options.

By understanding this process, students can confidently analyze and solve similar chemistry problems involving gas volumes and reaction stoichiometry.